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ARCH 331 Note Set 22.1 F2013abn1Reinforced Concrete Design

Notation:a = depth of the effective compressionblock in a concrete beamA = name for areaAg= gross area, equal to the total areaignoring any reinforcementAs= area of steel reinforcement inconcrete beam design

= area of steel compressionreinforcement in concrete beamdesignAst= area of steel reinforcement inconcrete column design

Av= area of concrete shear stirrupreinforcementACI = American Concrete Instituteb = width, often cross-sectionalbE= effective width of the flange of aconcrete T beam cross sectionbf= width of the flangebw= width of the stem (web) of aconcrete T beam cross sectioncc = shorthand for clear coverC = name for centroid= name for a compression forceCc= compressive force in thecompression steel in a doublyreinforced concrete beamCs= compressive force in the concreteof a doubly reinforced concretebeamd = effective depth from the top of areinforced concrete beam to thecentroid of the tensile steeld´ = effective depth from the top of areinforced concrete beam to thecentroid of the compression steeldb= bar diameter of a reinforcing barD = shorthand for dead loadDL = shorthand for dead loadE = modulus of elasticity or Young’smodulus= shorthand for earthquake loadEc= modulus of elasticity of concreteEs= modulus of elasticity of steelf = symbol for stressfc= compressive stress= concrete design compressive stressfpu= tensile strength of the prestressingreinforcementfs= stress in the steel reinforcement forconcrete design= compressive stress in thecompression reinforcement forconcrete beam designfy= yield stress or strengthF = shorthand for fluid loadFy= yield strengthG = relative stiffness of columns tobeams in a rigid connection, as is h = cross-section depthH = shorthand for lateral pressure loadhf= depth of a flange in a T sectionItransformed= moment of inertia of a multi-material section transformed to onematerialk = effective length factor for columns= length of beam in rigid joint= length of column in rigid jointld= development length for reinforcingsteel= development length for hooksln= clear span from face of support toface of support in concrete designL = name for length or span length, as isl= shorthand for live loadLr= shorthand for live roof loadLL = shorthand for live loadMn= nominal flexure strength with thesteel reinforcement at the yieldstress and concrete at the concretedesign strength for reinforcedconcrete beam designMu=

Structural design standards for reinforced concrete are established by the Building Code andCommentary (ACI 318-11) published by the American Concrete Institute International, and usesultimate strength design (also known as limit state design).

f’c= concrete compressive design strength at 28 days (units of psi when used in equations)

Materials

Concrete is a mixture of cement, coarse aggregate, fine aggregate, and water. The cementhydrates with the water to form a binder. The result is a hardened mass with “filler” and pores.There are various types of cement for low heat, rapid set, and other properties. Other minerals orcementitious materials (like fly ash) may be added.1cfcbalancedcARCH 331 Note Set 22.1 F2013abn3ASTM designations areType I: Ordinary portland cement (OPC)Type II: Low temperatureType III: High early strengthType IV: Low-heat of hydrationType V: Sulfate resistant

The proper proportions, by volume, of the mix constituentsdetermine strength, which is related to the water to cement ratio(w/c). It also determines other properties, such as workability offresh concrete. Admixtures, such as retardants, accelerators, orsuperplasticizers, which aid flow without adding more water, maybe added. Vibration may also be used to get the mix to flow into forms and fill completely.

Slump is the measurement of the height loss from a compacted cone of fresh concrete. It can bean indicator of the workability.

Proper mix design is necessary for durability. The pH of fresh cement is enough to preventreinforcing steel from oxidizing (rusting). If, however, cracks allow corrosive elements in waterto penetrate to the steel, a corrosion cell will be created, the steel will rust, expand and causefurther cracking. Adequate cover of the steel by the concrete is important.

Deformed reinforcing bars come in grades 40, 60 & 75 (for 40 ksi, 60 ksi and 75 ksi yieldstrengths). Sizes are given as # of 1/8” up to #8 bars. For #9 and larger, the number is a nominalsize (while the actual size is larger).

Reinforced concrete is a composite material, and the average density is considered to be 150 lb/ft3.It has the properties that it will creep (deformation with long term load) and shrink (a result ofhydration) that must be considered.

Construction

Because fresh concrete is a viscous suspension, it is cast or placed and not poured. Formworkmust be able to withstand the hydraulic pressure. Vibration may be used to get the mix to flowaround reinforcing bars or into tight locations, but excess vibration will cause segregation,honeycombing, and excessive bleed water which will reduce the water available for hydrationand the strength, subsequently.

After casting, the surface must be worked. Screeding removes the excess from the top of theforms and gets a rough level. Floating is the process of working the aggregate under the surfaceand to “float” some paste to the surface. Troweling takes place when the mix has hydrated to thepoint of supporting weight and the surface is smoothed further and consolidated. Curing isallowing the hydration process to proceed with adequate moisture. Black tarps and curingcompounds are commonly used. Finishing is the process of adding a texture, commonly byusing a broom, after the concrete has begun to set.ARCH 331 Note Set 22.1 F2013abn4

Behavior

Plane sections of composite materials can stillbe assumed to be plane (strain is linear), butthe stress distribution is not the same in bothmaterials because the modulus of elasticity isdifferent. (f=E)

In order to determine the stress, we can define nas the ratio of the elastic moduli:

n is used to transform the width of the second material such that it sees the equivalent elementstress.

Transformed Section y and I

In order to determine stresses in all types of material inthe beam, we transform the materials into a singlematerial, and calculate the location of the neutral axisand modulus of inertia for that material.

The ultimate strength design method is similar to LRFD. There is a nominal strength that isreduced by a factor  which must exceed the factored design stress. For beams, the concreteonly works in compression over a rectangular “stress” block above the n.a. from elasticcalculation, and the steel is exposed and reaches the yield stress, Fy

For stress analysis in reinforced concrete beams the steel is transformed to concrete any concrete in tension is assumed to becracked and to have no strength the steel can be in tension, and is placed in thebottom of a beam that has positive bendingmoment

ARCH 331 Note Set 22.1 F2013abn602 )xd(nAxbxsThe neutral axis is where there is no stress and no strain. The concrete above the n.a. is incompression. The concrete below the n.a. is considered ineffective. The steel below the n.a. isin tension.

Because the n.a. is defined by the moment areas, we can solve for x knowing that d is thedistance from the top of the concrete section to the centroid of the steel:

x can be solved for when the equation is rearranged into the generic format with a, b & c in thebinomial equation:02 cbxaxbyaacbbx242

T-sections

If the n.a. is above the bottom of a flange in a Tsection, x is found as for a rectangular section.

If the n.a. is below the bottom of a flange in a Tsection, x is found by including the flange and thestem of the web (bw) in the moment area calculation:

x = location to the neutral axisb = width of stress blockfy= steel yield strengthAs= area of steel reinforcementd = effective depth of section= depth to n.a. of reinforcementWith C=T, Asfy =0.85 f´cba so a can be determined withbffAacys85.0

Criteria for Beam Design

For flexure design:Mu Mn = 0.9 for flexure (when the section is tension controlled)so for design, Mucan be set to Mn=T(d-a/2) =  Asfy(d-a/2)

Reinforcement Ratio

The amount of steel reinforcement is limited. Too much reinforcement, or over-reinforcing willnot allow the steel to yield before the concrete crushes and there is a sudden failure. A beamwith the proper amount of steel to allow it to yield at failure is said to be under reinforced.The reinforcement ratio is just a fraction:bdAρs(or p) and must be less than a valuedetermined with a concrete strain of 0.003 and tensile strain of 0.004 (minimum). When thestrain in the reinforcement is 0.005 or greater, the section is tension controlled. (For smallerstrains the resistance factor reduces to 0.65 – see tied columns - because the stress is less than theyield stress in the steel.) Previous codes limited the amount to 0.75balancedwhere balancedwasdetermined from the amount of steel that would make the concrete start to crush at the exactsame time that the steel would yield based on strain.b

As

a/2

T

T

n.a.

C

C

x

a=

1x

0.85f’c

actual stress

Whitney stress block

d

h

ARCH 331 Note Set 22.1 F2013abn8Flexure Design of ReinforcementOne method is to “wisely” estimate a height of the stress block, a, and solve for As, and calculatea new value for a using Mu.

1. guess a (less than n.a.)2.ycsfbaf.A850

3. solve for a fromsetting Mu= Asfy(d-a/2):ysufAMda2

4. repeat from 2. until a found from step 3 matches a used in step 2.

Design Chart Method:1. calculate2bdMRnn

2. find curve for f’c

and fyto get 3. calculate Asand a, where:andbdAsbffAacys85.0

Any method can simplify the size of dusing h = 1.1d

Maximum ReinforcementBased on the limiting strain of0.005 in the steel, x(or c) = 0.375d so

)d.(a 37501to find As-max

(1is shown in the table above)

Minimum ReinforcementMinimum reinforcement is providedeven if the concrete can resist thetension. This is a means to controlcracking.

Minimum required:

but not less than:

wherecfis in psi. This can be translated to but not less than)db(ffAwycs3)db(fAwys200(tensile strain of 0.004)

ycff 3minyf200fromReinforced Concrete, 7th,

Wang, Salmon, Pincheira, Wiley & Sons, 2007

for which

is permitted to be 0.9

ARCH 331 Note Set 22.1 F2013abn9Cover for ReinforcementCover of concrete over/under the reinforcement must be provided to protect the steel fromcorrosion. For indoor exposure, 1.5 inch is typical for beams and columns, 0.75 inch is typicalfor slabs, and for concrete cast against soil, 3 inch minimum is required.

Bar SpacingMinimum bar spacings are specified to allow proper consolidation ofconcrete around the reinforcement. The minimum spacing is themaximum of 1 in, a bar diameter, or 1.33 times the maximum aggregate size.

T-beams and T-sections (pan joists)Beams cast with slabs have an effective width, bE,that sees compression stress in a wide flange beam orjoist in a slab system with positive bending.

For interior T-sections, bEis the smallest ofL/4, bw+ 16t, or center to center of beams

When the web is in tension the minimum reinforcement required is the same as for rectangularsections with the web width (bw) in place of b.When the flange is in tension (negative bending), theminimum reinforcement required is the greater value of orwherecfis in psi, bwis the beam width,and bfis the effective flange width

Compression Reinforcement

If a section is doubly reinforced, it means there is steel inthe beam seeing compression. The force in the compressionsteel that may not be yielding isCs= As´(f´s- 0.85f´c)

The total compression that balances the tension is now:T = Cc+ Cs. And the moment taken about the centroid ofthe compression stress is Mn= T(d-a/2)+Cs(a-d’)where As‘ is the area of compression reinforcement, and d’ is the effective depth to thecentroid of the compression reinforcement

Because the compression steel may not be yielding, the neutral axis x must be found from the forceequilibrium relationships, and the stress can be found based on strain to see if it has yielded.)(6dbffAwycs)(3dbffAfycsARCH 331 Note Set 22.1 F2013abn10Slabs

One way slabs can be designed as “one unit”-wide beams. Because they are thin, control ofdeflections is important, and minimum depthsare specified, as is minimum reinforcement forshrinkage and crack control when not inflexure. Reinforcement is commonly smalldiameter bars and welded wire fabric.Maximum spacing between bars is alsospecified for shrinkage and crack control asfive times the slab thickness not exceeding18”. For required flexure reinforcement thespacing limit is three times the slab thicknessnot exceeding 18”.

The maximum shear for design, Vuis the value at a distance of d from the face of the support.

Nominal Shear Strength

The shear force that can be resisted is the shear stress  cross section area:dbVwcc

The shear stress for beams (one way)ccf 2sodbfVwcc 2

where bw= the beam width or the minimum width of the stem. = 0.75 for shearOne-way joists are allowed an increase of 10% Vcif the joists are closely spaced.Stirrups are necessary for strength (as well as crack control):sdfAVyvsdbfwc8(max)where Av= area of all vertical legs of stirrups = spacing of stirrupsARCH 331 Note Set 22.1 F2013abn11d = effective depthARCH 331 Note Set 22.1 F2013abn12For shear design:

SCUVVV  

 = 0.75 for shear

Spacing RequirementsStirrups are required when Vuis greater than2Vc

Economical spacing of stirrups is considered to be greater than d/4. Commonspacings of d/4, d/3 and d/2 are used to determine the values of Vsat whichthe spacings can be increased.

This figure shows the size of Vnprovided by Vc+ Vs(long dashes) exceeds Vu/ in a step-wisefunction, while the spacing provided (short dashes) is at or less than the required s (limited by themaximum allowed). (Note that the maximum shear permitted from the stirrups isdbfwc8)The minimum recommended spacing for the first stirrup is 2 inches from the face of the support.sdfAVyvs ARCH 331 Note Set 22.1 F2013abn13Torsional Shear Reinforcement

On occasion beam members will see twist along theaxis caused by an eccentric shape supporting a load,like on an L-shaped spandrel (edge) beam. Thetorsion results in shearing stresses, and closedstirrups may be needed to resist the stress that theconcrete cannot resist.

Development Length for Reinforcement

Because the design is based on the reinforcement attaining the yield stress, the reinforcementneeds to be properly bonded to the concrete for a finite length (both sides) so it won’t slip. Thisis referred to as the development length, ld. Providing sufficient length to anchor bars that needto reach the yield stress near the end of connections are also specified by hook lengths. Detailingreinforcement is a tedious job. Splices are also necessary to extend the length of reinforcementthat come in standard lengths. The equations are not provided here.

Development Length in Tension

With the proper bar to bar spacing and cover, the common development length equations are:#6 bars and smaller:cybdfFdl25or 12 in. minimum#7 bars and larger:cybdfFdl20or 12 in. minimumDevelopment Length in Compression

ybcybdFd.fFd.l 00030020

Hook Bends and ExtensionsThe minimum hook length iscbdhfdl1200

ARCH 331 Note Set 22.1 F2013abn14Modulus of Elasticity & Deflection

Ecfor deflection calculations can be used with the transformed section modulus in the elasticrange. After that, the cracked section modulus is calculated and Ecis adjusted.

Code values:

ccfE 000,57(normal weight)cccfwE 335.1, wc= 90 lb/ft3- 160 lb/ft3

Deflections of beams and one-way slabs need not be computed if the overall member thicknessmeets the minimum specified by the code, and are shown in Table 9.5(a) (see Slabs).

Criteria for Flat Slab & Plate System Design

Systems with slabs and supporting beams, joists or columns typically have multiple bays. Thehorizontal elements can act as one-way or two-way systems. Most often the flexure resistingelements are continuous, having positive and negative bending moments. These moment andshear values can be found using beam tables, or from code specified approximate design factors.Flat slab two-way systems have drop panels (for shear), while flat plates do not.

Concrete columns rarely see only axial force and must be designed for the combined effects ofaxial load and bending moment. The interaction diagram shows the reduction in axial load acolumn can carry with a bending moment.

Monolithically cast frames withbeams and column elements will havemembers with shear, bending andaxial loads. Because the joints canrotate, the effective length must bedetermined from methods like thatpresented in the handout on RigidFrames. The charts for evaluating kfor non-sway and sway frames can befound in the ACI code.

Frame Columns

Because joints can rotate in frames, the effective length of the column in a frame is harder todetermine. The stiffness (EI/L) of each member in a joint determines how rigid or flexible it is.To find k, the relative stiffness, G or , must be found for both ends, plotted on the alignmentcharts, and connected by a line for braced and unbraced fames.

whereE = modulus of elasticity for a memberI = moment of inertia of for a memberlc= length of the column from center to centerlb= length of the beam from center to center

 For pinned connections we typically use a value of 10 for . For fixed connections we typically use a value of 1 for .

Find the design moment, Mu, from the factored load combination of 1.2D + 1.6L. It is good practice to guess a beam size toinclude self weight in the dead load, because “service” means dead load of everything except the beam itself.

or max-0.005if you want to use =0.9ARCH 331 Note Set 22.1 F2013abn20Example 5A simply supported beam 20 ft long carries a service dead load of 425 lb/ft (including self weight) and a live load of500 lb/ft. Design an appropriate beam (for flexure only). Use grade 40 steel and concrete strength of 5000 psi.

SOLUTION:

Find the design moment, Mu, from the factored load combination of 1.2D + 1.6L. If self weight is not included in the serviceloads, you need to guess a beam size to include self weight in the dead load, because “service” means dead load of everythingexcept the beam itself.

We can find Rn at the maximum reinforcement ratio for our materials, keeping in mind max at a strain = 0.005 is 0.0319 off of thechart at about 1070 psi, with max = 0.037. Let’s substitute b for a function of d:

Rn = 1070 psi =)12())(55.0(778,722ftinftlbddRearranging and solving for d = 11.4 inchesThat would make b a little over 6 inches, which is impractical. 10 in is commonly the smallest width.

So if h is commonly 1.5 to 2 times the width, b, h ranges from 14 to 20 inches. (10x1.5=15 and 10x2 = 20)

Now calculating an updated Rn =646.2psi)ftin(122625in)(10in)(11.ftlb72,778

 now is 0.020 (under the limit at 0.005 strain of 0.0319), so the estimated area required, As,can be found:

As = bd = (0.020)(10in)(11.625in) = 1.98 in2

The number of bars for this area can be found from handy charts.(Whether the number of bars actually fit for the width with cover and space between bars must also be considered. If you are atmax-0.005do not choose an area bigger than the maximum!)

We can’t just use max-.005. The way to reduce Rnis to increase b or d or both. Let’s try increasing h to 31 in., then Rn= 661 psiwith d = 28.625 in.. That puts us under max-0.005. We’d have to remember to keep UNDER the area of steel calculated, which ishard to do.From the chart,   0.013, less than the max-0.005of 0.0135, so the estimated area required, As,can be found:As= bd = (0.013)(15in)(29.625in) = 5.8 in2

The number of bars for this area can be found from handy charts. Our charts say there can be 3 – 6 bars that fit when ¾”aggregate is used. We’ll assume 1 inch spacing between bars. The actual limit is the maximum of 1 in, the bar diameter or 1.33times the maximum aggregate size.

Example 8Design a T-beam for a floor with a 4 in slab supported by 22-ft-span-length beams cast monolithically with the slab.The beams are 8 ft on center and have a web width of 12 in. and a total depth of 22 in.; f’c= 3000 psi and fy= 60 ksi.Service loads are 125 psf and 200 psf dead load which does not include the weight of the floor system.

ARCH 331 Note Set 22.1 F2013abn29Example 14Design the shear reinforcement for the simply supportedreinforced concrete beam shown with a dead load of 1.5 k/ftand a live load of 2.0 k/ft. Use 5000 psi concrete and Grade60 steel. Assume that the point of reaction is at the end of thebeam.

our maximum falls into the d/2 or 24”, so d/2 governs with 11.75 in Our 10” is ok.This spacing is valid until Vu = Vc and that happens at (64.9 k – 29.9 k)/0.451 k/in = 78 inWe can put the first stirrup at a minimum of 2 in from thesupport face, so we need 10” spaces for (78 – 2 - 6 in)/10 in =7 even (8 stirrups altogether ending at 78 in)After 78” we can change the spacing to the required (but notmore than the maximum of d/2 = 11.75 in  24in);s = Avfy/ 50bw= 0.22 in2(60,000 psi)/50 (12 in) = 22 inWe need to continue to 111 in, so (111 – 78 in)/ 11 in = 3even

8

-

#3 U stirrupsat 10 in

3

-

#3 U stirrups at11

in

2

in

Locating endpoints:

29.9 k = 64.9k–

0.451 k/in x (a)

a = 78 in

15 k = 64.9k–

0.451 k/in x (b)

b = 111 in.

15

29.9

78 in

111 in

ARCH 331 Note Set 22.1 F2013abn30Example 15

ARCH 331 Note Set 22.1 F2013abn31Example 15 (continued)

Example 16B

1.2

1.6

=1.2(93.8) + 1.6(250) = 112.6 + 400.0 = 516.2 psf (design load)

Because we are designing a slab segment that is 12 in. wide, the foregoing loading is the same as 512.6 lb/ft

or 0.513 kip/ft.

As-min= 0.12

in2/ft

No. 3 at 11 temperature reinforcement

No. 3 at 8

No. 3 at 8

No. 3 at 8

No. 3 at 9

No. 3 at 11

ARCH 331 Note Set 22.1 F2013abn32Example 16 (continued)

Similarly, the shears are determined using the ACI shear equations. In the end span at the face of the firstinterior support,

4.

Design the slab. Assume #4 bars for main steel with ¾ in. cover:d

= 5.5–

0.75–

½(0.5) = 4.5 in.

5.

Design the steel. (All moments must be considered.) For example, the negative moment in the end span at thefirst interior support:

kipsftun).)((.))((.bdMR 34054129010001220622

so



0.006

As

=bd

= 0.006(12)(4.5) = 0.325 in2

per ft. width of slabUse #4 at 7 in. (16.5 in. max. spacing)

The minimum reinforcement required for flexure is the same as the shrinkage and temperature steel.

(Verify the moment capacity is achieved:a

0.67 in. andMn

= 6.38 ft-kips > 6.20 ft-kips)

For grade 60 the minimum for shrinkage and temperature steel is:

As-min

=

0.0018bt

= 0.0018 (12)(5.5) = 0.12 in2

per ft. width of slabUse #3 at 11 in. (18 in. max spacing)

6.

Check the shearstrength.

lb.).)(()(.bdfVcc644365412300027502 

= 4.44 kips

Vu



Vc

Therefore the thickness is O.K.

7.

Development length for the flexure reinforcement is required. (Hooks are required at the spandrel beam.)

For example, #6 bars:

cybdfFdl25

or

12 in. minimum

With grade 40 steel and 3000 psi concrete:

inpsipsiinld9.21300025)000,40(86

(which is larger than 12 in.)

8.

Sketch:

(0.513)(11)2

= 4.43 ft-kips

(end span)

(0.513)(11)2

= 3.88 ft-kips

(interior span)

(0.513)(11)2

= 6.20 ft-kips

(end span-

first interior support)

(0.513)(11)2

= 5.64 ft-kips

(interior span–

both supports)

(0.513)(11)2

= 2.58 ft-kips

(end span–

exterior support)

1.15(0.513)

3.24 kips

(end span–

first interior support)

2.82 kips

=(0.513)

#3 at 11” o.c.

#4

at7” o.c.

#4

at8” o.c.

#4 at 12” o.c.

#4

at15” o.c.

#3 at 11” o.c.

temperature reinforcement

ARCH 331 Note Set 22.1 F2013abn33Example 17A building is supported on a grid of columns that is spaced at 30 ft on center in both the north-south and east-westdirections. Hollow core planks with a 2 in. topping span 30 ft in the east-west direction and are supported on precastL and inverted T beams. Size the hollow core planks assuming a live load of 100 lb/ft2. Choose the shallowestplank with the least reinforcement that will span the 30 ft while supporting the live load.

SOLUTION:

The shallowest that works is an 8 in. deep hollow core plank.The one with the least reinforcing has a strand pattern of 68-S, which contains 6 strands of diameter 8/16 in. = ½ in. The Sindicates that the strands are straight. The plank supports a superimposed service load of 124 lb/ft2at a span of 30 ft with anestimated camber at erection of 0.8 in. and an estimated long-time camber of 0.2 in.